The half life of `Te^(99)` is 6h. The activity of `Te^(99)` in a patient, 36 h after receiving an injection containing this isotope is `0.125 muci`. What was the minimum activity of the sample injected?

Answer

8

Explanation

Here, `T=6h, t=36h, A=0.125 muci, A_(0)=?`
As `A/(A_(0))=N/(N_(0))=(1/2)^(n)=(1/2)^(t//T)=(1/2)^(36//6)=1/64`
`:. A_(0)=64A=64xx0.125 muci =8muCi`